Question

1. equation: opens: vertex: axis of symmetry: p - value: focus: directrix: domain: range:

Explanation:

Step1: Determine the vertex

The vertex is the mid - point between the focus and the directrix. From the graph, the vertex is \((-2,0)\).

Step2: Find the value of \(p\)

The distance from the vertex to the focus (or directrix) is \(p\). The focus is at \((-2,0)\) and a point on the parabola is \((-4,5)\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) between \((-2,0)\) and \((-4,5)\) is not the right way. Since the parabola opens to the left (because the focus is to the left of the vertex). The standard form of a parabola that opens left is \((y - k)^2=4p(x - h)\), where \((h,k)\) is the vertex. The distance from the vertex \((-2,0)\) to the focus \((-4,0)\) is \(p=- 2\) (negative because it opens left).

Step3: Write the equation

Using the standard form \((y - k)^2 = 4p(x - h)\) with \((h,k)=(-2,0)\) and \(p=-2\), we get \((y-0)^2=4\times(-2)(x + 2)\), so \(y^{2}=-8(x + 2)\)

Step4: Axis of symmetry

For a parabola of the form \((y - k)^2=4p(x - h)\), the axis of symmetry is \(y = k\). Here \(k = 0\), so \(y = 0\)

Step5: Focus

From the graph, the focus is \((-4,0)\)

Step6: Directrix

The directrix is \(x=h - p\). Since \(h=-2\) and \(p=-2\), \(x=-2-(-2)=0\)

Step7: Domain and Range

For the parabola \(y^{2}=-8(x + 2)\), solving for \(x\) gives \(x=-\frac{y^{2}}{8}-2\). The domain is \(x\leqslant-2\) (because \(p<0\)) and the range is all real numbers, \(y\in(-\infty,\infty)\)

Answer:

Equation: \(y^{2}=-8(x + 2)\)
Opens: Left
Vertex: \((-2,0)\)
Axis of Symmetry: \(y = 0\)
\(p\) - value: \(-2\)
Focus: \((-4,0)\)
Directrix: \(x = 0\)
Domain: \(x\leqslant-2\)
Range: \(-\infty